Dissociation equation for ch3cooh

Asked by maham237 @ 09/01/2021 in Chemistry viewed by 264 People

Acetic acid (CH3COOH) is a weak acid that partially dissociates in water. Given the reaction CH3COOH(aq) ↔ CH3COO−(aq) + H+(aq) and Kc = 1.8 x 10−5, if the concentration of the acetic acid is 0.016 M and the concentration of the acetate ion is 0.92 M, what is the [H+]?

Answered by maham237 @ 09/01/2021

Answer:

The correct answer is: 1.035 x 10⁻³ M

Explanation:

The dissociation equilibrium for acetic acid (CH₃COOH) is the following:

CH₃COOH(aq) ↔ CH₃COO⁻(aq) + H⁺(aq) Kc = 1.8 x 10⁻⁵

The expression for the equilibrium constant (Kc) is the ratio of concentrations of products over reactants. The products are acetate ion (CH₃COO⁻) and hydrogen ion (H⁺) while the reactant is acetic acid (CH₃COOH):

$Kc=\frac{[CH_{3} COO^{-} ][H^{+} ]}{[CH_{3} COOH]}= 1.8 x 10^{-5}$

Given: [CH₃COOH]= 0.016 M and [CH₃COO⁻]= 0.92 M, we replace the concentrations in the equilibrium expression and we calculate [H⁺]:

$\frac{(0.016 M)[H^{+} ]}{(0.92M)}= 1.8 x 10^{-5}$

⇒[H⁺]= (1.8 x 10⁻⁵)(0.92 M)/(0.016 M)= 1.035 x 10⁻³ M

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